Chapter 12: Q 31. (page 830)

Use mathematical induction to prove that $a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d] = n a + d \frac{n (n - 1)}{2}$

Short Answer

Expert verified

The given statement is shown.

Step by step solution

Step 1. Given information

The statement is $a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d] = n a + d \frac{n (n - 1)}{2}$

Step 2. Prove the given statement.

The first term in the given series is $a$

Put $n = 1$ in $n a + d \frac{n (n - 1)}{2}$ to check if the statement is true for $n = 1$ .

Assume that the statement is true for $n = k$

Thus, $a + (a + d) + (a + 2 a) + \dots \dots + [a + (k - 1) d] = k a + d \frac{k (k - 1)}{2}$ .

Find the sum for $k + 1$ terms and check whether the sum equals

width="207">(k+1)a+d(k+1)((k+1)-1)2

$\begin{array}{rcl} a + (a + d) + (a + 2 d) + \dots \dots \dots \dots + [a + (k - 1) d] + [a + ((k + 1) - 1) d] & = & = k a + d \frac{k (k - 1)}{2} + a + k d \\ = & a (k + 1) + \frac{d}{2} [k^{2} - k + 2 k] \\ = & a (k + 1) + \frac{d}{2} [(k + 1) ((k + 1) - 1)] \end{array}$

Thus, the statement is true for k + 1.

Since both the conditions of the Principle of Mathematical Induction are satisfied, the statement is true for all natural numbers.

Therefore, $a + (a + a) + (a + 2 a) + \dots \dots + [a + (n - 1) d] = n a + d \frac{n (n - 1)}{2}$ .

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Use mathematical induction to prove that $a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d] = n a + d \frac{n (n - 1)}{2}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Prove the given statement.

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Use mathematical induction to prove thata+(a+d)+(a+2d)+⋯+[a+(n-1)d]=na+dn(n-1)2

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Step by step solution

Step 1. Given information

Step 2. Prove the given statement.

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Use mathematical induction to prove that $a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d] = n a + d \frac{n (n - 1)}{2}$